Question

Consider a fish tank that initially contains 300 liters of water with 100 grams of salt dissolved in it. Water containing 4 grams of salt per liter is allowed to run into the tank at a rate of 2 liters per minute. The thoroughly stirred mixture in the tank is also allowed to drain out at a rate of 3 liters per minute.

(a) Find a differential equation for the amount of salt x(t) in the tank at time t.
(b) Solve the DE in (a).

Answer 1

Explanation:

given that a fish tank that initially contains 300 liters of water with 100 grams of salt dissolved in it. Water containing 4 grams of salt per liter is allowed to run into the tank at a rate of 2 liters per minute.

The thoroughly stirred mixture in the tank is also allowed to drain out at a rate of 3 liters per minute.

Let X(t) be the amount of salt in the tank at time t

X(0) = 100 gms.

Volume of tank at time t = initial volume+t(incoming rate-outgoing rate)

= 300+(2-3)t= 300 -t

Rate of change of X would be=incoming salt rate-outgoing salt rate

`= 4(2) -(3X(t))/(300-t)`,X(0) = 100 gm

i.e.
`(dx)/(dt) = 8-(3X(t))/(300-t)`

`(dx)/(dt) +(3X(t))/(300-t)= 8`

this is linear DE

`e^{\int\limits {(3)/(300-t) } \, dt } =e^(-3ln (300-t)) \\=(1)/((300-t)^3)`

Solution woul dbe

`X*(1)/((300-t)^3) =\int\limits{(8)/((300-t)^3) } \, dt\\=(4)/((300-t)^2) +C\\X(t) = 4(300-t)+C(300-t)^3`

Use initial condition to find t

X(0) = 100 = 1200+300^3
`X(t) = 4(300-t)-4.074 (300-t)^3`C

C = -4.074(10^-5)